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Exams and Problem Solutions
 
--Matters and Properties of Matters Exams and Problem Solutions
 
--Atomic Structure Exams and Problem Solutions
 
--Periodic Table Exams and Problem Solutions
 
--The Mole Concept Exams and Problem Solutions
 
--Gases Exams and Problem Solutions
 
--Chemical Reactions Exams and Problem Solutions
 
--Nuclear Chemistry (Radioactivity) Exams and Problem Solutions
 
--Nuclear Chemistry (Radioactivity) Exam1 and Problem Solutions
 
--Nuclear Chemistry (Radioactivity) Exam2 and Problem Solutions
 
--Solutions Exams and Problem Solutions
 
--Acids and Bases Exams and Problem Solutions
 
--Thermochemistry Exams and Problem Solutions
 
--Rates of Reaction Exams and Problem Solutions
 
--Chemical Equilibrium Exams and Problem Solutions
 
--Chemical Bonds Exams and Problem Solutions
 
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Nuclear Chemistry (Radioactivity) Exam1 and Problem Solutions


Nuclear Chemistry (Radioactivity) Exam1 and Problem Solutions

1. Find whether 90231Th is stable or not.

Solution:

n+p=Mass Number

90+n=231

n=141

where n is number of neutrons and p is number of protons.Thus ratio between n and p is;

n0/p+=141/90=1,56

Since ratio is greater than 1, 90231Th has unstable nucleus and it is radioactive element.


2. A, B, C and D elements form compounds AC, A2D and BD. If AC and A2D are radioactive and BD is not radioactive compound, find whether the following compounds are radioactive or not.

I. A2

II. A2C

III. C2D

IV. BC

Solution:

If a compound is radioactive, at least one of the elements of this compound must be radioactive. Since BD is not radioactive, B and D are not radioactive elements. If AC and A2D are radioactive then A must be radioactive element C can be radioactive or not we can not say anything about it.

A2 and A2C are radioactive compounds because of radioactive element A but we can not say whether C2D and BC are radioactive or not.


3. Find number of protons and mass number of Y in given reaction below.

92234X + β- + α → Y + γ + 2β+

Solution:

Number of protons in left side of reaction is;

92 +(-1) + 2 =93

Thus, number of protons in right side of reaction must be 89.

Y+1.(0) + 2.(+1) = 93

Y=91

Y=91 number of protons

Mass number of reactants must be equal to mass numbers of products.

234+4=238 mass number of reactants

Y+ 1.(0) + 2.(0) =238

Y =238

Y=238 mass number of Y

91238Y


4. Find X and Y in given reactions.

I. 1938K → 1838Ar + X

II. 80197Hg + Y → 79197Au

Solution:

I. 1938K → 1838Ar + abX

mass number and atomic numbers must be equal;

38=38+b

b=0

19=18+a

a=1 thus, +10X or +10β

II. 80197Hg + cdY → 79197Au

80 + c = 79

c=-1

197 + d = 197

d=0 So, Y = -10β


5. Which ones of the following statements are true for atom having following reaction in its nucleus?

11p → 01n + +10β

I. Its mass number increases by 1.

II. Its isotope is formed.

III. Its netron number decreases by 1.

IV. Its atomic number decreases by 1.

V. Its number of protons increases by 1.

Solution:

In given reaction one proton is converted into one neutron. Thus, atomic number decreases by 1. IV is true.


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